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20101ee2_1_Practice_set1_solution

20101ee2_1_Practice_set1_solution - EE 2 Solutions to...

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EE 2 Solutions to Practice Problem Set 1 [1] Note: Z(E) is the same as D(E) in class notes, i.e. the density of states function Z(E)dE = no of available electron states in the energy range E to E+dE. Therefore, no of (available) electron states in the energy range E 1 to E 2 is given by 2 1 ) ( E E dE E Z Remember from class notes (pg 21, eq 36) that dE E h m V dE E Z 2 / 1 3 2 / 3 ) 2 ( 4 ) ( π = Note that the volume of the rectangular box is V = 1x2x3 = 6cm 3 or 6x10 -6 m 3 . Therefore, the required number of states is ) ( 3 2 . ) 2 ( 4 3 2 . ) 2 ( 4 ) 2 ( 4 2 / 3 1 2 / 3 2 3 2 / 3 2 / 3 3 2 / 3 2 / 1 3 2 / 3 2 1 2 1 E E h m V E h m V dE E h m V E E E E = = = π π π Substituting for the various values, we have No of states with energy in the range from 1eV to 1.2eV is ( ) ( ) 21 2 / 3 19 2 / 3 19 3 34 2 / 3 31 6 10 57 . 8 10 602 . 1 1 10 602 . 1 2 . 1 3 2 . ) 10 626 . 6 ( ) 10 11 . 9 2 ( 10 6 4 × = × × × × × × × × × = π [2] At any temperature T (in Kelvin), the probability that an available state at energy
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